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Homework Portfolio

1.

- 1.71. Our proof of the Cauchy-Schwarz inequality, Theorem 1.13, used that when \(U\) is a unit vector, \[0 \leq ||V−(U·V)U||^2 = ||V||2 −(U·V)^2\]. Therefore if \(U\) is a unit vector and equality holds, then \(V = (U · V)U\). Show that equality occurs in the Cauchy Schwarz inequality for two arbitrary vectors \(V\) and \(W\) only if one of the vectors is a multiple (perhaps zero) of the other vector.

Answer: In the first case, when \(W=0\), the \(W\) is a multiple of \(V\); In the second case, when \(W\) is nonzero, then consider the unit vector \(U = \frac{W}{||W||}\). Then, by the result in the question, it follows that \(V = (U · V)U\). Therefore:

\[V = (U · V)U = (\frac{W}{||W||}· V)·\frac{W}{||W||}= (\frac{W·V}{||W||^2})·W\] As \((\frac{W·V}{||W||^2})\) is a constant, so \(V\) is a multiple of \(W\).

2.

-2.19. Suppose C is an n by n matrix with orthonormal columns. Use Theorem 2.2 to show that \[||CX|| \leq \sqrt{n} ||X||\] Use the Pythagorean theorem and the result of Problem 2.17 to show that in fact \[||CX|| = ||X||\] for such a matrix.

Answer: (1) First we calculate \(C\). Let \(C_j\) denote the \(j\)th column of C. Since C has orthonormal columns, each \(C_j\) has norm 1. Then \[||C|| = \sqrt{\mathop{\sum_{i=1}^n\sum_{j=1}^n}C_{ij}^2}\] \[=\sqrt{\mathop{\sum_{j=1}^n(\sum_{i=1}^n}C_{ij}^2)}\] \[=\sqrt{\sum_{j=1}^n|| C_j||^2}\] Now by Theorem 2.2, \[||CX|| \leq ||C||||X|| = \sqrt{n}||X||\] as desired. (2)By Problem 2.17, \(CX=x_1C_1 +x_2C_2 +\dots+x_nC_n\). To find the norm of RHS, we need to apply the Pythagorean theorem (we need that C has orthonormal columns) to get \[||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2 = ||x_1 C_1||^2 + ||x_2 C_2||^2 + \dots + ||x_n C_n||^2\] Now we can put the pieces together: \[||CX||^2 = ||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2\] \[=||x_1 C_1||^2 + ||x_2 C_2||^2+\dots+ ||x_n C_n||^n\] \[=x_1^2||C_1||^2 + x_2^2|| C_2||^2+\dots+ x_n ^2||C_n||^n\] \[=x_1^2 + x_2^2+\dots+ x_n ^2\] \[= ||X||^2\] Since norms are nonnegative, we can conclude that \(||CX|| = ||X||\).

3.

-2.44. Use the Cauchy-Schwarz inequality \[|A · B| \leq||A||||B||\]

to prove: (a) the function \(f (X) = C · X \)is uniformly continuous,

(b) the function \(g(X, Y) = X · Y\) is continuous.

Answer: (a) In case 1, If \(C=0\) then \(f(X)=0\) for all \(X\),so \(|f(X)−f(Y)|=0<ε\) for all \(ε,X,Y\).

In case 2, where \(C=(c1,c2,...,cn)\neq(0,0,...,0)\).

By the definition of f and properties of the dot product, \(|f(X)− f(Y)|=|C·Y−C·Y|=|C·(X−Y)|\).

By the Cauchy-Schwartz inequality we get \[|f(X)− f(Y)|=|C·(X−Y)|\leq||C||||X−Y||\] Let \(ε>0\) and take \(δ=\frac {ε}{||C||}\)

If \(||X−Y||<δ= \frac {ε}{||C||}\) Then we have \(|f(X)− f(Y)|\leq||C||||X−Y||<ε\) for all \(X\) and \(Y\) in \(Rn\). Therefore for any C, f is uniformly continuous.

(b) Fix \(V (A,B)\) in \(R^{2n}\), to show that \(g(V)= g(X,Y)= XY\) is continuous at \((a,b)\)

Given \(\epsilon>0\), we need to find \(\delta>o\) If \(||U-V||=\sqrt {(X-A)^2+(Y-B)^2}<\delta\)

then \(||g(U)-g(V)||=|XY-AB|<\epsilon\)

By the triangle inequality, we know that \[|XY-AB|= |[(X-A)+A][(Y-B)+B]-AB|\] \[=|(X-A)(Y-B)+B(X-A)+A(Y-B)|\] \[\leq||X-A||+||B|| ||X-A||+||A|| ||Y-B||\] If \(\sqrt {(X-A)^2+(Y-B)^2}<\delta\) Then \(||X-A||+||B|| ||X-A||+||A|| ||Y-B||\) \[\leq (1+||A||+||B||)\sqrt {(X-A)^2+(Y-B)^2}\] \[\leq(1+||A||+||B||)\epsilon\] \[\leq(1+||A||+||B||) \frac {\delta}{1+||A||+||B||}\] \[=\delta\] So, given \(\epsilon>0\), set \(\delta= min(1,\frac {\delta}{1+||A||+||B||})\) we have \(\sqrt {(X-A)^2+(Y-B)^2}<\delta\) Therefore, for any \((A,B)\) in \(R^{2n}\), \(g(X, Y) = X · Y\) is continuous.

4.

-2.45. In the triangle inequality \(||A + B|| \leq ||A|| + ||B||\) put \(A = X − Y\) and \(B = Y\). Deduce \(||Y|| − ||X|| \leq ||Y − X||\). Show that if two points are within one unit distance of each other, then the difference of their norms is less than or equal to one.

Answer: Let \(A\) and \(B \)be in \(R_n\). Apply the triangle inequality we have \[||A + B|| \leq ||A|| + ||B||.\] Let\(X=A+B\), \(Y=B\),so \(A=X−Y\). so we have \[||X|| \leq ||X − Y|| + ||Y||, ||X|| − ||Y|| \leq ||X − Y||.\]

Exchange the symbol \(X\) and \(Y\) we get \[||Y|| − ||X|| \leq ||Y − X||.\] So when \(X\) and \(Y\) are within one unit of distance of each other which means \( ||Y − X|| \leq 1\), by the inequality we can conclude \[||Y|| − ||X|| \leq ||Y − X||\leq 1\]

5.

-3.9. Suppose a function \(F\) from \(R_n\) to \(R_m\) is differentiable at \(A\). Justify the following statements that prove \[L_AH = DF(A)H,\] that is, the linear function LA in Definition 3.4 is given by the matrix of partial derivatives\( DF(A).\)

(a) There is a matrix\( C\) such that \(L_A(H) = CH\) for all \(H\).

(b) Denote by \(C_i\) the \(i-th\) row of \(C\). The fraction \[\frac{||F(A+H)−F(A)−L_A(H)||}{||H||} = \frac{||F(A+H)−F(A)−CH||}{||H||}\] tends to zero as \(||H||\) tends to zero if and only if each component \[\frac {f_i(A+H)− f_i(A)−C_iH}{ ||H||}\] tends to zero as \(||H||\) tends to zero.

(c) Set \(H = h e_j\) in the \(i-th \)component of the numerator to show that the partial derivative \(f_{i,x_j} (A)\) exists and is equal to the \((i, j)\) entry of \(C\).

Answer: (a) \(F\) from \(R_n\) to \(R_m\) is differentiable at \(A\). By the definition of differentiability there is a linear function \(L_A(H)\) such that \[\frac{||F(A+H)−F(A)−L_A(H)||}{||H||}\]

tends to \(0\) as \(||H||\) tends to zero.

By Theorem 2.1, every linear function from \(R_n\) to \(R_m\) can be written as \(LA(H)=CH\) for all \(H\), where \(C\) is some \(m × n\) matrix.

(b) The absolute value of each component of a vector is less than or equal to the norm of the vector so for each H and i we have \[0 \leq| f_i(A + H) − f_i(A) − C_i · H| \leq ||F(A + H) − F(A) − CH||\] where \(Ci\) is the \(i-th\) row of \(C\).

Since \[\frac{||F(A+H)−F(A)=LA(H)||}{||H||}\] tends to 0,

by the squeeze theorem, both\(\frac{ | f_i(A+H)− f_i(A)−C_i·H|}{||H|| }\) and\(\frac{ f_i(A+H)− f_i(A)−C_i·H}{||H|| }\)

(c) Let \(H = he_j = (0,0,...,1,0,...,0)\), the 1 in the\( j-th\) place. By part (b), tend to zero as ||H|| tends to zero. \[\lim_{\||H||=|h|\to\ 0 } \frac{f_i(A+he_j)− f_i(A)−hC_i ·e_j}{|h|} =0\]

Therefore \[\lim_{h\to\ 0 } \frac{f_i(A+he_j)− f_i(A)−hC_i ·e_j}{h} =0\]

Since \(C_i · e_j = c_{i j}\) and by the definition of partial derivatives,we have \[\lim_{h\to\ 0 } \frac{f_i(A+he_j)− f_i(A)}{h} = \frac {\partial f_i}{\partial x_j} (A)\]

So we can conclude that \(c_{i j}= c (A) \)

6.

6.14. Justify the following items which prove:

If f is continuous on \(R_2\) and \(\int_{R}f dA = 0\) for all smoothly bounded sets \(R\), then \(\) is identically zero.

(a) If \(f(a,b)=p>0\) then there is a disc \(D\) of radius \(r>0\) centered at \((a,b) \)in which \(f(x,y)> \frac {1}{2}p\)

(b) If f is continuous and \(f(x,y) ≥ p_1 > 0\) on a disk \(R\) then \(\int_{R}f dA \geq p_1(Area(R))\). \(\int_{R}f dA = 0\)for all smoothly bounded regions \(R\), then \(f\) cannot be positive at any point. (d) \(f\) is not negative at any point either. (e) \(f = 0\) at all points.

Answer: First, we can assume that f is continuous on R2 and \(\int_{R}f dA = 0\) for all smoothly bounded sets \(R\)

(a) Because f is continuous at \((a, b)\), by the definition of continuity, there is \(r > 0\) such that for all \((x, y)\) such that\( ||(x, y) − (a, b)|| < r\), we have \(|f(x, y) − f(a, b)| < p/2\).Then we assume p > 0, so \(p − p/2 < f(x,y) < p + p/2\) In particular, \(f (x, y) > p/2\)

(b) As \(R\) is bounded, the closure of \(R\) is closed and bounded. So we can apply the extreme value theorem which means \(f\) is bounded on the closure of \(R\). In particular, \(f\) is bounded on \(R\). \(f\) is also integrable on \(R\); in fact \(\int_{R}f dA = 0\). Apply the lower bound property, \(\int_{R}f dA \geq p_1(Area(R))\) holds.

(c) Suppose \(f\) is positive at \((a, b)\).

From \((a)\), there is a disc \(R \)of nonzero radius on which \( f(x,y) > f(a,b)/2 > 0\).

From (b), \(\int_{R}f dA \geq (f(a,b)/2)·area(R) > 0\) But we assumed that \(\int_{R}f dA = 0\) for all smoothly bounded sets \(R\), it comes to a contradiction. Therefore \(f\) cannot be positive at any point.

(d) As we know that \(−f\) is continuous, and that for all smoothly bounded regions \(R\), by linearity, we have \(−fdA=− fdA=−0=0\) . From(c),we know that \(-f\) cannot be positive at any point. Thus, we conclude that f cannot be negative at any point.

(e) Therefore, for any \((a,b)\), \(f(a,b)\) is defined and is neither positive nor negative, so it must be \(0\).

7.

-6.44. Justify the following steps to prove that if \(f\) is integrable on \(R_2\) and \(g\) is a continuous function with \(0 \leq g \leq f\) then \(g\) is integrable on \(R_2\).

(a) \(\int_{D(n)} g dA\) exsits

(b) \(0 \leq \int_{D(n)} g dA \leq \int_{D(n)} f dA\)

(c) The numbers \(\int_{D(n)} g dA\) are an increasing sequence bounded above.

(d) \(\lim_{n\to\infty} \int_{D(n)} g dA\) exsits

Answer:

Check \(D : D= R^2\) unbounded, g 0, continuous, so we need to prove \(\lim_{n\to\infty} \int_{D(n)} g dA\) exsits.

(a) \(g \geq 0\) is continuous on \(R_2\) and \(D(n)\) is bounded for each \(n\) so \(g\) is integrable over \(D(n)\)

(b) By theorem 6.9 \(L area(D) \leq I (f, D)\) and the fact 0g, we know that \( 0 area(D)\leq \int_{D(n)} g dA\) so if \( 0 \leq f(x,y)−g(x,y)\) then \[0=0 area(D)\leq \int_{D(n)} f(x,y)-g(x,y) dA \leq \int_{D(n)} f dA - \int_{D(n)} g dA\] Therefore, \(0 \leq \int_{D(n)} g dA \leq \int_{D(n)} f dA\)

(c) Let \(C_n =\int_{D(n)} g dA\). Because \(g \geq 0\), \(D(n) \leq D(n+1)\). Then \(C_1, C_2, C_3...C_n\) is an increasing sequence. Since \(0 \leq \int_{D(n)} g dA \leq \int_{D(n)} f dA\) and \[\lim_{n\to\infty} \int_{D(n)} f dA = \int_{R ^2} f dA\] exists, We got \[\int_{D(n)} g dA \leq \int_{R ^2} f dA\]

(e) By the Monotone Convergence Theorem for sequences, \(\int_{D(n)} g dA\) increasing and bounded above is convergent so \(\lim_{n\to\infty} \int_{D(n)} g dA = \lim_{n\to\infty} C_n\) exists

8.

6.50. Justify steps (a)–(d) to prove that if a continuous function \(f\) is integrable on an unbounded set \(D\) then \(\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA\)

(a)\(\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} \left|f \right|dA\)

(b)\(\int_ {D} (-f) dA \leq \int_ {D} \left|f \right|dA\)

(c)\(- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA\)

(d)\(\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA\)

(a) By Definition 6.9, if \(f\) is continuous and integrable on an unbounded set \(D\), then \(| f |\) is integrable on \(D\). Rewrite \(f(x,y) = f_+(x,y)− f_−(x,y)\) where \(f_+(x,y) = f(x,y)\) if \(f(x,y) \geq 0\) and 0 otherwise, and \(f_−(x,y) = −f(x,y)\)if \(f(x,y)\leq 0\) and \(0\) otherwise. So, by the definition of \(\int_ {D} f dA\), \[\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA\] Since \(\int_ {D} f_- dA\) is nonnegtive \[\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA\] Since \(f_+ ≥ 0\) and \(f_− ≥ 0\) are integrable over \(D\) \[\int_ {D(n)} f_+ dA + \int_ {D(n)} f_- dA =\int_ {D(n)} (f_+ + f_-) dA\] By the properties of limits of increasing sequence D(n), we know \(\int_ {D(n)} (f_+ + f_-) dA\) converges so \[\int_ {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} (f_+ + f_-) dA\] By the equation \(f(x,y) = f_+(x,y)− f_−(x,y)\), we got \[\int_ {D} f dA \leq \int_ {D} \left|f \right|dA\]

(b) In the same way, we apply (a) to the functions − f to get \[\int_ {D} -f dA \leq \int_ {D} \left|-f \right|dA= \int_ {D} \left|f \right|dA\]

(c)By the properties of limits and the equation \(\int_ {D(n)} -f dA =\)_ D(n) f dA \(, we get\)\(- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA\)\( (d)If \)ba\( and\) \(−b\leq a\) then\(|b|\leq a.\) From (a), we got \[\int_ {D} f dA \leq \int_ {D} \left|f \right|dA\], From (b) and (c), we got \[- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA\] Therefore, we can conclude that \[\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA\]

9.

-4.21. Find the point on the plane \[z = x − 2y + 3\] that is closest to the origin, by finding where the square of the distance between \((0,0)\) and a point \((x,y)\) of the plane is at a minimum. Use the matrix of second partial derivatives to show that the point is a local minimum.

Let \[D=d^2 = f(x,y)= x^2+y^2+(x-2y+3)^2\], to find the local extrema we let \[\triangledown f = (4x−4y+6,−4x+10y−12)=0\] at \((− 0.5,1).\) so \[H(-0.5,1)= \left[ \begin{array}{ c c } 4 & -4 \\ -4 & 10 \end{array} \right]\] Because 4 > 0 and (4)(10) − (−4)2 = 24 > 0. So by the Theorem 4.3, it is positive definite. By theorem 4.8, If \(\triangledown f (A) = 0\) and the Hessian matrix \([f_{x_i x_j} (A])\) is positive definite at \(A\), then \(f (A)\) is a local minimum. Therefore, \(f\) has a local minimum at point \((-0.5,1)\)

10.

-7.32. Let \(S\) be the unit sphere centered at the origin in \(R^3\). Evaluate the following items, using as little calculation as possible

(a)\(\int_{S} 1 d\sigma\)

(b)\(\int_{S} ||X||^2 d\sigma\)

(c) Verify that \(\int_{S} x_1^2 d\sigma = \int_{S} x_2^2 d\sigma =\int_{S} x_3^2 d\sigma\) using either a symmetric argument or parametrizations. Can you do this without evaluating them?

(d) Use the result of parts (b) and (c) to deduce the value of \(\int_{S} x_1^2 d\sigma\)

Answer:

(a) In geometry, \(\int_{S} 1 d\sigma\) means the area of the unit sphere in \(R ^3\) So \(\int_{S} 1 d\sigma= \pi·1^3 =4\pi\)

(b) For all X S we have \(||X||^2 =1\), therefore \(\int_{S} ||X||^2 d\sigma=\int_{S} 1 d\sigma=4\pi\)

(c) Rotation by /2 about the \(x_3\)-axis corresponds to some transformation on the domain of the parametrization of \(S\). We know that \(x_1\) comes to the same position as \(x_2\), Therefore \(\int_{S} x_1^2 d\sigma = \int_{S} x_2^2 d\sigma\) In the same way, make a rotation by \(\pi/2\) about the \(x_2\) , we got \(\int_{S} x_1^2 d\sigma = \int_{S} x_3^2 d\sigma\) Therefore, \(\int_{S} x_1^2 d\sigma = \int_{S} x_2^2 d\sigma =\int_{S} x_3^2 d\sigma\)

(d) By the definition of norm \(||X||\), we know that \( ||X||=x_1^2 +x_2^2 +x_3^2 \) So, \[\int_{S} ||X||^2 d\sigma= \int_{S} x_1^2 +x_2^2 +x_3^2 d\sigma= 3\int_{S} x_1^2 d\sigma = 4\pi\] Therefore,\[\int_{S} x_1^2 d\sigma =\frac{1}{3}\int_{S} ||X||^2 d\sigma = \frac{4\pi}{3}\]

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